Practice Exercises for Chapter 4

Practice Exercises for Chapter 4 Pauline Chow
Use the reduction of order method to find the second linearly independent solution of the differential equation.

b( x)
dx
e a( x)
Formula: a( x) y "( x)  b( x) y '( x)  c( x)  0; y2  y1 ( x)  2
dx
y1 ( x)
1.
y” + 9y = 0; y1 = sin 3x
3.
y” + 2y’ + y = 0; y1 = e –x
Solve the higher order ODE:
5.
y”’ – 6y” + 12y’ – 8y = 0
8.
y(5) – 2y(4) + 17y”’ = 0
6.
9.
4x2 y” + y = 0: y1 = x1/2
x2y” – 3xy’ + 5y = 0; y1 = x2 cos (ln x)
2.
4.
y”’ + 3y” + 3y’ + y = 0
3y”’ – 14y” + 20y’ – 8y = 0
Solve the higher order Cauchy-Euler ODE:
11.
x2 y” – 7xy’ + 41y = 0
13
x3 y’’’– 2x2y” +4xy’ – 4y = 0
15.
x3 y’’’ + xy’ – y = 0
17.
x4 y(4) + 6x3 y”’ + 9x2 y” + 3x y’ + y = 0
7.
10.
y”’ + 5y” = 0
y(4) – 9y = 0
x3 y’’’– 2x2y” – 2xy’ + 8y = 0
x3 y’’’ + 6y = 0
x4 y(4) + 6x3 y”’ = 0
x3 y’’’+ 5x2y” + 7xy’ + 8y = 0
12.
14.
16.
18.
Use undetermined coefficients method and the annihilator method, if applicable, to solve.
19.
y”’ – 6y” = 3 – cos x
20.
y”’ – 2y” – 4y’ + 8y = 6e2x
x
21.
y”’ – 3y” + 3y’ – y = x – 4e
22.
y”’ – y” – 4y’ + 4y = 8 – 3ex + 4e2x
–2x
23.
y”’ + 8y = 24x – 16 + 24e
24.
2y” + 5y’ – 3y = –7e–3x – 42 sin 3x + 30 cos 3x
2x
2
2
25.
y” – 2y’ + 5y = 15e + 25x
26.
x y” + 5xy’ – 21y = –54x2 – 48x – 84, y(1) = 13, y’(1) = 7
2
2
2
27.
x y” – xy’ + y = 2x ln x + 10x
Use parameter of variations method to solve.
28.
x3 y’’’ – 3x2y” + 6xy’ – 6y = 4 + ln x
30.
y”’ + 4y’ = sec 2x
32.
2y”’ – 6y” = x2
34.
x2 y” – 4xy’ + 6y = 1/x
36.
y” – 6y’ + 9y = 27x
38.
x2 y” + 4xy’ – 10y = 28 lnx
29.
31.
33.
35.
37.
y”’ + y’ = tan x
y”’ – 2y” – y’ + 2y = e3x
y” – 4y’ + 4y = (12x2 – 6x) e2x
y” + y = cot x
y” – 4y’ + 13y = 36e2x
Use any method to solve:
39.
y(4) + 2y” + y = (x – 1)2
40.
41.
y”’ – 2y” + y’ = 2 – 24ex + 40e5x, y(0) = ½, y’(0) = 5/2, y”(0) = –9/2
y(4) – y” = 4x + 2xe–x
Solutions:
1. y2 = –(1/3) cos 3x 2. y2 = x 1/2 ln x 3. y2 = xe–x 4. y2 = x2 sin (ln x)
5. (r – 2)3 = 0; y = C1 e2x + C2 xe2x + C3 x2 e2x
6. (r + 1)3 = 0; y = C1 e–x + C2 xe–x + C3 x2 e–x
2
–5x
8. r3(r2 – 2r + 17) = 0; y = C1 + C2 x + C3 x2 + C4 ex cos 4x + C5 ex sin 4x
7. r ( r + 5) = 0; y = C1 + C2 x + C3 e
2
2x
2x
2/3x
9. (r – 2) (3r – 2) = 0; y = C1 e + C2 xe + C3 e
10. (r2 + 3)(r2 – 3) = 0; y  C1 cos 3 x  C2 sin 3 x  C3 e 3 x  C4 e 3 x 11. y = C1 x4 cos (5 ln x) + C2 x4 sin (5 ln x)
12. r3 – 5r2 + 2r + 8 = (r + 1)(r – 2)(r – 4) = 0; y = C1 x-1 + C2 x2 + C3 x4
13. r3 – 5r2 + 8r – 4 = (r – 1)(r – 2)2 = 0; y = C1 x + C2 x2 + C3 x2 ln x
14. r3 – 3r2 + 2r + 6 = (r + 1)(r2 – 4r + 6) = 0; y  C1 x 2 cos
3


2 ln x  C2 x 2 sin


2 ln x  C3 x 1
2
16. r(r – 1)(r – 2)(r + 3) = 0; y = C1 + C2 x + C3 x2 + C4 x–3
15. (r – 1) = 0; y = C1 x + C2 x ln x + C3 x (ln x)
4
2
2
2
17. r + 2r + 1 = (r + 1) = 0; y = C1 cos (ln x) + C2 sin (ln x) + C3 (ln x) cos (ln x) + C4 (ln x) sin (ln x)
18. r3 + 2r2 + 4r + 8 = (r + 2)(r2 + 4) = 0; y = C1x–2 + C2 cos (2 lnx) + C3 sin (2 lnx)
1
6
1
3
19. y  C1 C2 x  C3 e6 x  x 2  cos x  sin x
20. y  C1e 2 x  C2 e 2 x  C3 xe 2 x  x 2 e 2 x
4
37
37
4
2 3 x
x
x
2 x
x
2 x
2x
21. y  C1e  C2 xe  C3 x e  3  x  x e
22, y  C1e  C2 e  C3 e  2  xe x  xe 2 x
3
23. y  C1e x cos 3 x  C2 e x sin 3 x  C3 e 2 x  3 x  2  2e 2 x
2
x
x
2x
2
25. y  C1e cos 2 x  C2 e sin 2 x  3e  5 x  4 x 
5
24. y  C1e
x/2
 C2 e
3 x

xe
3 x
26. y = –2x3 + 2x–7x + 6x2 + 3x + 4
 2 sin 3 x
35 1
 ln x
36 6
29. y = C1 + C2 cos x + C3 sin x – ln | cos x| | + 1 – sin x ln | sec x + tan x |
1
1
1
30. y  C1 C2 cos x  C3 sin x  ln | sec x  tan x |  x cos 2 x  sin 2 x ln | cos 2 x |
8
4
4
1
1 4 1 3 1 2 1
1
32. y  C1 x  C2 x  C3 e3 x 
x 
x 
x  x
31. W  6e 2 x ; y  C1e  x  C2 e x  C3 e 2 x  e3 x
3
72
54
54
81
243
1 1
2x
2x
4
3
2x
2
3
34. y  C1 x  C2 x  x
35. y = C1 cos x + C2 sin x – sin x ln | csc x + cot x |
33. y  C1e  C2 xe  ( x  x )e
12
14
5
2 14
36. y = C1 e3x + C2 xe3x + 3x + 2 37. y = C1 e2x cos 3x + C2 e2x sin 3x + 4e2x 38. y  C1 x  C2 x  ln x 
25
5
39. y = C1 cos x + C2 sin x + C3 x cos x + C4 x sin x + x2 – 2x – 3
2
5
1
1
40. y  C1  C2 x  C3 e x  C4 e x  x3  xe  x  x 2 e x
41. y  11  11e x  9 xe x  2 x  12 x 2 e x  e5 x
2
3
2
2
More questions:
28. y  C1 x C2 x 2  C3 x 3 
27. y = C1x + C2 x lnx + 2x2 lnx + 6x2
Find a differential operator that annihilates the given function.
2.
6e2x cos 3x + 5e2x sin 3x
1.
e4x + 4xe4x + 3x – 9
–x
4.
x3 + 4x – 10
3.
xe cos 5x
Solutions:
1. D2 (D – 4)2 2. D2 – 4D + 13
3. (D2 + 2D + 26)2
4. D4
Given the homogenous solutions, find the particular solution using an appropriate method.
1. x4 y(4) + 2x3 y’” – 3x2 y” + 19xy’ + 13y = x4 /16; yh = C1 x3 cos (2ln x) + C2 x3 sin (2 ln x) + C3 x –1 + C4 x –1 ln x
2. x2 y” – 9xy’ + 25y = 0; yh = C1 x5 + C2 x5 ln x
3. x2 y” – xy’ – 3y = 0; yh = C1 x3 + C2 x –1
4. y(4) – 4y’” + 2y” + 20y’ + 13y = 90e2x; yh = C1 e3x cos 2x + C2 e3x sin 2x + C3 e –x + C4 xe –x
5. ; y(4) – 4y’” + 28y” + 4y’ – 29y = 21ex cos x + 52ex sin x; yh= C1 e2x cos 5x + C2 e2x sin 5x + C3 e –x + C4 ex
6. y” – 10y’ + 25y = 82 sin 2x; yh = C1 e5x + C2 xe5x;
7. y” – 2y’ – 3y = 6xe2x; yh = C1 e3x + C2 e –x
8. y” – 8y’ + 20y = 60x3 + 8x2 – 6x – 8; yh = C1 e4x cos 2x + C2 e4x sin 2x
x
9. y’” – 2y” + 4y’ = 36x2 + 4x ; yh  C1e cos
Solutions:
5. yp = –ex cos x
4. yp = 2e2x
3
2
9. yp = 3x + 5x + x


x
3 x  C2 e sin
6. yp = 2 sin 2x


3 x  C3
7. yp = –2xe2x + (4/3)e2x
Use the Elimination Method to find a general solution for the given linear system.
1.
x’ = 2x – y
y’ = x
2.
x’ = –y + t
y’ = x – t
3.
x’ = 4x + 7y
y’ = x – 2y
4.
x’ – 4y = 1
x + y’ = 2
5.
x’ + y’ = et
–x” + x’ + x + y = 0
6.
x’ = –5x – y
y’ = 4x – y
x(1) = 0, y(1) = 1
8. yp = 3x3 + 4x2 + 2x
7.
x’ = y – 1
y’ = –3x + 2y
x(0) = 0, y(0) = 0
8.
D2 x – Dy = 1
(D + 3)x + (D + 3)y = 2
9.
(D2 – 1)x – y = 0
(D – 1)x + Dy = 0
10.
Dx + D2 y = e3t
(D + 1)x + (D – 1)y = 4e3t
11.
x’ – 6y = 0
x – y’ + z = 0
x + y – z’ = 0
12.
x’ = –x + z
y’ = –y + z
z’ = –x + y
13.
2Dx + (D – 1)y = t
Dx + Dy = t2
14.
Dx – 2Dy = t2
(D – 1)x – 2(D + 1)y = 1
Solutions:
1. x = C1 et + C2 tet, y = (C1 – C2)et + C2 tet
2. x = C1 cos t + C2 sin t + t + 1, y = C1 sin t – C2 cos t – 1 + t
1
3. x = C1 e5t + C2 e–3t, y  C1e5t  C2 e3t
7
1
1
1
4. x = C1 cos 2t + C2 sin 2t + 2, y   C1 sin 2t  C2 cos 2t 
2
2
4
5. x = C1 + C2 t + C3 et + tet, y = (–C1 + C2) – C2 t + (1 – C3) et – tet
6. x = (1 – t)e3 – 3t, y = (2t – 1)e3 – 3t
1
2 t
2
e sin 2t  , y  1  et cos 2t
7. x   et cos 2t 
3
3
3
2
8. x = C1 + C2 e–3t + C3 e–t + t, y   C1  3C2 e 3t  C3 e t  t
3
3
3
1
3
1
3
t  C3 e 1/ 2t sin
t , y   (3C2  3C3 )e 1/ 2t cos
t  (3 C3  3C2 )e 1/ 2t sin
t
2
2
2
2
2
2
17
4
10. x  C1 + C2 sin t  C3 cos t  e3t , y  C1 + C2 cos t  C3 sin t  e3t
15
15
1
1
1
5
1
1
11. x = C1 e–t + C2 e3t + C3 e–2t, y   C1e  t  C2 e3t  C3 e 2t , z   C1e  t  C2 e3t  C3 e 2t
6
2
3
6
2
3
12. x = C3 e–t + C1 + C2 te–t, y = (C3 – C2)e–t + C1 + C2 te–t, z = C1 + C2 e–t
9. x  C1et  C2 e 1/ 2t cos
13. x = (1/3)t3 – C1 e–t + C2 – 2t2 + 5t, y = C1 e–t + 2t2 – 5t + 5
14. x 
1 2 1 3
1
1
1
t  t  C1 , y  t 2  t 3  (1  C1 )
2
6
4
12
2